Class 12 Physics Alternating Current Power associated with Inductor

Power associated with Inductor

  • Voltage and current associated with the inductor V = Vm sinωt  is  I = Im sin(ωt –(∏/2)) respectively.
  • Instantaneous power p = VI = Vm Im sinωt sin(ωt –(∏/2))
  • = - Vm Im sinωt cos ωt  ( using  formula (sin(A –(∏/2)) = – cosA))
  • After simplifying we get,
  • p = - ((Vm Im sin2ωt)/(2))
  • Average Power PL = (1/T) 0T∫ p dt
    • =(-1/T) 0T∫ (Vm Im/2) sin2 ωt dt
    • = - ( Vm Im/2T) 0T∫ sin2 ωt dt
    • After simplifying  we will get ,
  • PL  =  0.
  • Therefore average power supplied to an inductor over one complete cycle is 0.

Problem:- A pure inductor of 25.0 mH is connected to a source of 220 V. Find the inductive reactance and rms current in the circuit if the frequency of the source is 50 Hz.

Answer:-

The inductive reactance,

XL = 2πν L = (2x3.14x50x25x10-3) W

= 7.85Ω

The rms current in the circuit is:- I =(V/XL) =(220V)/(7.85Ω) =28A

 

Problem:-A coil of inductance 0.50 H and resistance 100 Ω is connected to a 240 V,

50 Hz ac supply. (a) What is the maximum current in the coil? (b) What is the time lag between the voltage maximum and the current maximum?

Answer:-

Inductance of the inductor, L = 0.50 H

Resistance of the resistor, R = 100 Ω

Potential of the supply voltage, V = 240 V

Frequency of the supply, ν = 50 Hz

(a) Peak voltage is given as: V0=√2V

=√2 x 240 =339.41V

Angular frequency of the supply, ω = 2 πν

= 2π × 50 = 100 π rad/s

Maximum current in the circuit is given as:

I0= (V0)/( √(R)2 + ω2L2)

=(339.41)/((100)2+(100π)2+(0.50)2) = 1.82A

(b) Equation for voltage is given as:

V = V0 cos ωt

Equation for current is given as:

I = I0 cos (ωt − Φ)

Where,

Φ = Phase difference between voltage and current At time, t = 0.

V = V0(voltage is maximum)

For ωt − Φ = 0 i.e., at time ,

I = I0 (current is maximum)

Hence, the time lag between maximum voltage and maximum current is .( Φ/ω)

Now, phase angle Φ is given by the relation,

tan Φ = (ωL)/(R)

= (2 π x 50x 0.5)/(100)

=1.57

Φ = 57.50 = (57.5 π)/(180)rad

ωt = (57.5 π)/(180)

t =(57.5)/(180 x 2 π x 50)

=3.19x10-3 s

=3.2ms

Hence, the time lag between maximum voltage and maximum current is 3.2 ms.

Problem:- Obtain the answers (a) to (b) in previous problem if the circuit is connected to a high frequency supply (240 V, 10 kHz). Hence, explain the statement that at very high frequency, an inductor in a circuit nearly amounts to an open circuit. How does an inductor behave in a dc circuit after the steady state?

Answer:-

Inductance of the inductor, L = 0.5 Hz

Resistance of the resistor, R = 100 Ω

Potential of the supply voltages, V = 240 V

Frequency of the supply, ν = 10 kHz = 104Hz

Angular frequency, ω = 2πν= 2π × 104rad/s

  • Peak voltage, V0=√2V = 240√2 V

Maximum current , I0= (V0)/( √(R)2 + ω2L2)

=(240√2) / ((100)2 + (2 π x 104)2 x(0.50)2)

=1.1 x 10-2 A

 (b) For phase difference Φ, we have the relation:

tan Φ = (ωL)/(R)

= (2 π x 104 x 0.5)/(100) =100 π

Φ =89.820 = (89.82 π)/(180) rad

ω t = (89.82 π)/(180)

t= (89.82 π)/(180x2 πx104)

=25 μ s

It can be observed that I0 is very small in this case. Hence, at high

frequencies, the inductor amounts to an open circuit.

In a dc circuit, after a steady state is achieved, ω = 0. Hence, inductor L behaves like a pure conducting object.

(Magnetisation and demagnetisation of a magnet)

  • Current i through the coil entering at point A increase from zero to a maximum value. Flux lines are set up i.e., the core gets magnetised. With the polarity shown voltage and current are both positive. So their product p is positive.

      ENERGY IS ABSORBED FROM THE SOURCE.

  • Current in the coil is still positive but is decreasing. The core gets demagnetised and the net flux becomes zero at the end of a half cycle. The voltage v is negative (since di/dt is negative). The product of voltage and current is negative, and ENERGY IS BEING RETURNED TO SOURCE.

(One complete cycle of voltage/current . It is clear  that the current lags  the voltage.)

  • Current i becomes negative i.e., it enters at B and comes out of A. Since the direction of current has changed, the polarity of the magnet changes. The current and voltage are both negative. So their product p is positive. ENERGY IS ABSORBED.

  • Current i decreases and reaches its zero value at 4 when core is demagnetised and flux is zero. The voltage is positive but the current is negative. The power is, therefore, negative. ENERGY ABSORBED DURING THE ¼ CYCLE 2-3 IS RETURNED TO THE SOURCE.

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